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Oscillatory behaviour of a higher-order dynamic equation

Journal of Inequalities and Applications20132013:52

https://doi.org/10.1186/1029-242X-2013-52

  • Received: 26 September 2012
  • Accepted: 30 January 2013
  • Published:

Abstract

In this paper we are concerned with the oscillation of solutions of a certain more general higher-order nonlinear neutral-type functional dynamic equation with oscillating coefficients. We obtain some sufficient criteria for oscillatory behaviour of its solutions.

MSC:34N05.

Keywords

  • time scale
  • higher-order nonlinear neutral dynamic equation
  • oscillating coefficient

1 Introduction

The calculus on time scales has been introduced in order to unify the theories of continuous and discrete processes and in order to extend those theories to a more general class of the so-called dynamic equations. In recent years there has been much research activity concerning the oscillation and non-oscillation of solutions of neutral dynamic equations on time scales.

In this paper we consider the higher-order nonlinear dynamic equation
[ y ( t ) + P ( t ) y ( τ ( t ) ) ] Δ n + i = 1 m Q i ( t ) f i ( y ( ϕ i ( t ) ) ) = 0 ,
(1.1)

where n 2 , P ( t ) , Q i ( t ) C r d [ t 0 , ) T for i = 1 , 2 , , m ; P ( t ) is an oscillating function ( P ( t ) : T R ), Q i ( t ) are positive real-valued functions for i = 1 , 2 , , m ; ϕ i ( t ) C r d [ t 0 , ) T , ϕ i Δ ( t ) > 0 , the variable delays τ , ϕ i : [ t 0 , ) T T with τ ( t ) , ϕ i ( t ) < t for all t [ t 0 , ) T , ϕ i ( t ) as t for i = 1 , 2 , , m ; τ ( t ) as t ; f i ( u ) C ( R , R ) are nondecreasing functions, u f i ( u ) > 0 for u 0 and i = 1 , 2 , , m .

The purpose of the paper is to study oscillatory behaviour of solutions of equation (1.1). For the sake of convenience, the function z ( t ) is defined by
z ( t ) = y ( t ) + P ( t ) y ( τ ( t ) ) .
(1.2)

2 Basic definitions and some auxiliary lemmas

A time scale T is an arbitrary nonempty closed subset of the real numbers . For t T , we define the forward jump operator σ : T T by
σ ( t ) : = inf { s T : s > t }
while the backward jump operator ρ : T T is defined by
ρ ( t ) : = sup { s T : s < t } .
If σ ( t ) > t , we say that t is right-scattered, while if ρ ( t ) < t , we say that t is left-scattered. Also, if σ ( t ) = t , then t is called right-dense, and if ρ ( t ) = t , then t is called left-dense. The graininess function μ : T [ 0 , ) is defined by
μ ( t ) : = σ ( t ) t .

We introduce the set T κ which is derived from the time scale T as follows. If T has left-scattered maximum m, then T κ = T { m } , otherwise T κ = T .

Definition 1 [1]

The function f : T R is called rd-continuous provided it is continuous at right-dense points in T and its left-sided limits exist (finite) at left-dense points in T .

Theorem 1 [1]

Assume that ν : T R is strictly increasing and T ˜ : = ν ( T ) is a time scale. Let w : T ˜ R . If ν Δ ( t ) and w Δ ˜ ( ν ( t ) ) exist for t T κ , then
( w ν ) Δ = ( w Δ ˜ ν ) ν Δ ,

where we denote the derivative on T ˜ by Δ ˜ .

Definition 2 [1]

Let f : T R be a function. If there exists a function F : T R such that F Δ ( t ) = f ( t ) for all t T κ , then F is said to be an antiderivative of f. We define the Cauchy integral by
a b f ( τ ) Δ ( τ ) = F ( b ) F ( a ) for  a , b T .

Theorem 2 [2]

Let u and v be continuous functions on [ a , b ] that are Δ-differentiable on [ a , b ) . If u Δ and v Δ are integrable from a to b, then
a b u Δ ( t ) v ( t ) Δ ( t ) + a b u σ ( t ) v Δ ( t ) Δ ( t ) = u ( b ) v ( b ) u ( a ) v ( a ) .
Let T ˜ = T { sup T } { inf T } . If T ˜ , we call ∞ left-dense, and −∞ is called right-dense provided T ˜ . For any left-dense t 0 T ˜ and any ε > 0 , the set
L ε ( t 0 ) = { t T : 0 < t 0 t < ε }

is nonempty, and so is L ε ( ) = { t T : t > 1 ε } if T ˜ .

Lemma 1 [3]

Let n N and f be n-times differentiable on T . Assume T ˜ . Suppose there exists ε > 0 such that
f ( t ) > 0 , sgn ( f Δ n ( t ) ) s { 1 , + 1 } for all t L ε ( )
and f Δ n ( t ) 0 on L δ ( ) for any δ > 0 . Then there exists p [ 0 , n ] N 0 such that n + p is even for s = 1 and odd for s = 1 with
{ ( 1 ) p + j f Δ j ( t ) > 0 for all t L ε ( ) , j [ p , n 1 ] N 0 , f Δ j ( t ) > 0 for all t L δ j ( ) ( with δ j ( 0 , ε ) ) , j [ 1 , p 1 ] N 0 .

Lemma 2 [3]

Let f be n-times differentiable on T κ n , t T , and α T κ n . Then with the functions h k defined as h n ( t , s ) = ( 1 ) n g n ( s , t ) ,
h 0 ( r , s ) 1 and h k + 1 ( r , s ) = s r h k ( τ , s ) Δ s for k N 0 ,
we have
f ( t ) = k = 0 n 1 h k ( t , α ) f Δ k ( α ) + α ρ n 1 ( t ) h n 1 ( t , σ ( τ ) ) f Δ n ( τ ) Δ τ .

Lemma 3 [3]

Let f be n-times differentiable on T κ n and m N with m < n . Then we have, for all α T κ n 1 + m and t T κ m ,
f Δ m ( t ) = k = 0 n m 1 h k ( t , α ) f Δ k + m ( α ) + α ρ n m 1 ( t ) h n m 1 ( t , σ ( τ ) ) f Δ n ( τ ) Δ τ .

Lemma 4 [3]

Suppose f is n-times differentiable and g k , 0 k n 1 , are differentiable at t T κ n with
g k + 1 Δ ( t ) = g k ( σ ( t ) ) for all 0 k n 2 .
Then we have
[ k = 0 n 1 ( 1 ) k f Δ k g k ] Δ = f g 0 Δ + ( 1 ) n 1 f Δ n g n 1 σ .

3 Main results

Lemma 5 Let f be n-times differentiable on T κ n . If f Δ > 0 , then for every λ, 0 < λ < 1 , we have
f ( t ) λ ( 1 ) n 1 g n 1 ( σ ( T ) , t ) f Δ n 1 ( t ) .
(3.1)
Proof Let p, 0 p n 1 , be the integer assigned to the function f as in Lemma 1. Because of f Δ > 0 , we always have p > 0 . Furthermore, let T T be assigned to f by Lemma 1. Then, by using the Taylor formula on time scales, for every ρ n 1 ( t ) T , we obtain
f ( t ) T ρ n 1 ( t ) ( 1 ) n 1 g n 1 ( σ ( τ ) , t ) f Δ n ( τ ) Δ τ .
(3.2)
By using Theorem 2 and (3.2), we have
f ( t ) ( 1 ) n 1 g n 1 ( σ ( t ) , t ) f Δ n 1 ( t ) T ρ n 1 ( t ) ( 1 ) n 1 g n 1 ( σ ( τ ) , t ) f Δ n 1 ( τ ) Δ τ .
Since f is n-times differentiable on T κ n and m N with m < n , we have with n and f substituted by n m and f Δ m , respectively
f Δ m ( t ) T ρ n m 1 ( t ) ( 1 ) n m 1 g n m 1 ( σ ( τ ) , t ) f Δ n ( τ ) Δ τ .
Also, for every ρ n 1 ( t ) , s with ρ n 1 ( t ) T and T s t , we have
f Δ m ( s ) ( 1 ) n m 1 g n m 1 ( σ ( T ) , t ) f Δ n ( t ) .
This is obvious for m = n 1 and, when m < n 1 , it can be derived by applying the Taylor formula. Thus, for all t T , we get
f ( t ) ( 1 ) n 1 g n 1 ( σ ( T ) , t ) f Δ n 1 ( t )

and therefore the proof of the lemma can be immediately completed. □

The result of Lemma 5 is an extension of studies in [4] and [5]. In order that the reader sees how the results in [4] (1.8.14) and [5] (Lemma 2) follow from (3.1), it is at this point only necessary to know that in the case T = Z , we have ρ ( t ) = t 1 , σ ( t ) = t + 1 and
g n 1 ( σ ( T ) , t ) = ( t T 1 ) ( n 1 ) ( n 1 ) ! ,
then we get the inequality in [4]
u ( t ) 1 ( n 1 ) ! ( n n 1 ) ( n 1 ) Δ n 1 u ( 2 n m 1 n ) ;
and in the case T = R , we have ρ ( t ) = σ ( t ) = t and
g n 1 ( σ ( T ) , t ) = ( t T ) ( n 1 ) ( n 1 ) ! ,
then we get the inequality in [5]
u ( t ) ϑ ( n 1 ) ! ( t ) n 1 u n 1 ( t ) .

For the cases T = Z and T = R , some sufficient criterias for oscillatory behaviour of the solutions of the equation (1.1) were obtained by Bolat and Akın in [6] and [7], respectively. Furthermore, there might be other time scales that we cannot appreciate at this time due to our current lack of ‘real-world’ examples.

Theorem 3 Assume that n is odd and

(C1) lim t P ( t ) = 0 ,

(C2) t 0 s n 1 i = 1 m Q i ( s ) Δ s = .

Then every bounded solution of equation (1.1) is either oscillatory or tends to zero as t .

Proof Assume that equation (1.1) has a bounded non-oscillatory solution y ( t ) . Without loss of generality, assume that y ( t ) is eventually positive (the proof is similar when y ( t ) is eventually negative). That is, y ( t ) > 0 , y ( τ ( t ) ) > 0 and y ( ϕ i ( t ) ) > 0 for t t 1 t 0 and i = 1 , 2 , , m . Assume further that y ( t ) does not tend to zero as t . By (1.1), (1.2), we have for t t 1
z Δ n ( t ) = i = 1 m Q i ( t ) f i ( y ( ϕ i ( t ) ) ) < 0 .
(3.3)

It follows that z Δ j ( t ) ( j [ 0 , n 1 ] N 0 ) is strictly monotone and eventually of constant sign. Since P ( t ) is an oscillatory function, there exists a t 2 t 1 such that if t t 2 , then z ( t ) > 0 . Since y ( t ) is bounded, by virtue of (C1) and (1.2), there is a t 3 t 2 such that z ( t ) is also bounded for t t 3 . Because n is odd and z ( t ) is bounded, by Lemma 1, when p = 0 (otherwise z ( t ) is not bounded), there exists t 4 t 3 such that for t t 4 we have ( 1 ) j z Δ j ( t ) > 0 , j [ 0 , n 1 ] N 0 .

In particular, since z Δ ( t ) < 0 for t t 4 , z ( t ) is decreasing. Since z ( t ) is bounded, we write lim t z ( t ) = L ( < L < ). Assume that 0 L < . Let L > 0 . Then there exists a constant c > 0 and a t 5 t 4 such that z ( t ) > c > 0 for t t 5 . Since y ( t ) is bounded, lim t P ( t ) y ( τ ( t ) ) = 0 by (C1). Therefore, there exists a constant c 1 > 0 and a t 6 t 5 such that y ( t ) = z ( t ) P ( t ) y ( τ ( t ) ) > c 1 > 0 for t t 6 . So that we can find a t 7 with t 7 t 6 such that y ( ϕ i ( t ) ) > c 1 > 0 for t t 7 . From (3.3) we have
z Δ n ( t ) = i = 1 m Q i ( t ) f i ( c 1 ) < 0
(3.4)
for t t 7 . If we multiply (3.4) by t n 1 and integrate it from t 7 to t, we obtain
F ( t ) F ( t 7 ) f ( c 1 ) t 7 t i = 1 m Q i ( s ) s n 1 Δ s ,
(3.5)
where
F ( t ) = i = 1 n 1 ( 1 ) i + 1 ( t n 1 ) Δ i z Δ n i ( σ i ( t ) )
and
σ i ( t ) = σ ( σ i 1 ( t ) ) .
Since ( 1 ) k z Δ k ( t ) > 0 for k = 0 , 1 , 2 , , n 1 and t t 4 , we have F ( t ) > 0 for t t 7 . From (3.5) we have
F ( t 7 ) f ( c 1 ) t 7 t i = 1 m Q i ( s ) s n 1 Δ s .
By (C2) we obtain
F ( t 7 ) f ( c 1 ) t 7 t i = 1 m Q i ( s ) s n 1 Δ s =
as t . This is a contradiction. So, L > 0 is impossible. Therefore, L = 0 is the only possible case. That is, lim t z ( t ) = 0 . Since y ( t ) is bounded, by (C1) we obtain
lim t y ( t ) = lim t z ( t ) lim t P ( t ) y ( t ) = 0

from (1.2).

Now let us consider the case of y ( t ) < 0 for t t 1 . By (1.1) and (1.2),
z Δ n ( t ) = i = 1 m Q i ( t ) f i ( y ( ϕ i ( t ) ) ) > 0

for t t 1 . That is, z Δ n > 0 . It follows that z Δ j ( t ) ( j [ 0 , n 1 ] N 0 ) is strictly monotone and eventually of constant sign. Since P ( t ) is an oscillatory function, there exists a t 2 t 1 such that if t t 2 , then z ( t ) < 0 . Since y ( t ) is bounded, by (C1) and (1.2), there is a t 3 t 2 such that z ( t ) is also bounded for t t 3 . Assume that x ( t ) = z ( t ) . Then x Δ n ( t ) = z Δ n ( t ) . Therefore, x ( t ) > 0 and x Δ n ( t ) < 0 for t t 3 . Hence, we observe that x ( t ) is bounded. Since n is odd, by Lemma 1, there exists a t 4 t 3 and p = 0 (otherwise x ( t ) is not bounded) such that ( 1 ) j x Δ j ( t ) > 0 , j [ 0 , n 1 ] N 0 and t t 4 . That is, ( 1 ) j z Δ j ( t ) < 0 , j [ 0 , n 1 ] N 0 and t t 4 . In particular, for t t 4 we have z Δ ( t ) > 0 . Therefore, z ( t ) is increasing. So, we can assume that lim t z ( t ) = L ( < L 0 ). As in the proof of y ( t ) > 0 , we may prove that L = 0 . As for the rest, it is similar to the case of y ( t ) > 0 . That is, lim t y ( t ) = 0 . This contradicts our assumption. Hence the proof is completed. □

Theorem 4 Assume that n is even and (C 1) holds. If the following condition is satisfied:

(C3) There is a function φ ( t ) such that φ ( t ) C r d 1 [ t 0 , ) T . Moreover,
lim t sup t 0 t φ ( s ) i = 1 m Q i ( s ) Δ s =
and
lim t sup t 10 t [ φ Δ ( s ) ] 2 φ ( s ) g n 2 σ ( σ ( ϕ i ( s ) ) , ϕ i ( s ) ) Δ s <

for φ ( t ) and i = 1 , 2 , , m . Then every bounded solution of equation (1.1) is oscillatory.

Proof Assume that equation (1.1) has a bounded non-oscillatory solution y ( t ) . Without loss of generality, assume that y ( t ) is eventually positive (the proof is similar when y ( t ) is eventually negative). That is, y ( t ) > 0 , y ( τ ( t ) ) > 0 and y ( ϕ i ( t ) ) > 0 for t t 1 t 0 . By (1.1) and (1.2), we have (3.3) for t t 1 . Then z Δ n ( t ) < 0 . It follows that z Δ j ( t ) ( j [ 0 , n 1 ] N 0 ) is strictly monotone and eventually of constant sign. Since P ( t ) is an oscillatory function, there exists a t 2 t 1 such that for t t 2 , we have z ( t ) > 0 . Since y ( t ) is bounded, by (C1) and (1.2), there is a t 3 t 2 , such that z ( t ) is also bounded for t t 3 . Because n is even, by Lemma 1 when p = 1 (otherwise z ( t ) is not bounded), there exists t 4 t 3 such that for t t 4 we have
( 1 ) j + 1 z Δ j ( t ) > 0 , j [ 0 , n 1 ] N 0 .
(3.6)
In particular, since z Δ ( t ) > 0 for t t 4 , z ( t ) is increasing. Since y ( t ) is bounded,
lim t P ( t ) y ( τ ( t ) ) = 0
by (C1). Let δ > 1 ; i.e., there exists a t 5 t 4 such that by (1.2)
y ( t ) = z ( t ) P ( t ) y ( τ ( t ) ) > 1 δ z ( t ) > 0
for t t 5 . We may find a t 6 t 5 such that for t t 6 and i = 1 , 2 , , m ,
y ( ϕ i ( t ) ) > 1 δ z ( ϕ i ( t ) ) > 0 .
(3.7)
From (3.3), (3.7) and the properties of f, we have
z Δ n ( t ) i = 1 m Q i ( t ) f i ( 1 δ z ( ϕ i ( t ) ) ) = i = 1 m Q i ( t ) f i ( 1 δ z ( ϕ i ( t ) ) ) z ( ϕ i ( t ) ) z ( ϕ i ( t ) )
(3.8)
for t t 6 . Since z ( t ) > 0 is bounded and increasing, lim t z ( t ) = L ( 0 < L < ). By the continuity of f, we have
lim t f i ( 1 δ z ( ϕ i ( t ) ) ) z ( ϕ i ( t ) ) = f i ( L δ ) L > 0 .
Then there is a t 7 t 6 such that for t t 7 , i = 1 , 2 , , m , we have
f i ( 1 δ z ( ϕ i ( t ) ) ) z ( ϕ i ( t ) ) f i ( L δ ) 2 L = α > 0 .
(3.9)
By (3.8), (3.9),
z Δ n ( t ) α i = 1 m Q i ( t ) z ( ϕ i ( t ) ) for  t t 7 .
(3.10)
Set
w ( t ) = z Δ n 1 ( t ) z ( 1 δ ϕ i ( t ) ) .
(3.11)
We know from (3.6) that there is a t 8 t 7 such that for a sufficiently large t t 8 , w ( t ) > 0 . Therefore, Δ-derivating (3.11) we obtain
w Δ ( t ) = z Δ n ( t ) z ( δ 1 ϕ i ( t ) ) + z Δ n 1 ( σ ( t ) ) ( 1 z ( δ 1 ϕ i ( t ) ) ) Δ = z Δ n ( t ) z ( δ 1 ϕ i ( t ) ) δ 1 ϕ i Δ ( t ) z Δ n 1 ( σ ( t ) ) z Δ ( δ 1 ϕ i ( t ) ) z ( δ 1 ϕ i ( t ) ) z ( δ 1 ϕ i ( σ ( t ) ) )
(3.12)
z Δ n ( t ) z ( δ 1 ϕ i ( t ) ) δ 1 ϕ i Δ ( t ) z Δ n 1 ( σ ( t ) ) z Δ ( δ 1 ϕ i ( t ) ) z 2 ( δ 1 ϕ i ( σ ( t ) ) ) = z Δ n ( t ) z ( δ 1 ϕ i ( t ) ) δ 1 w σ ( t ) ϕ i Δ ( t ) z Δ ( δ 1 ϕ i ( t ) ) z ( δ 1 ϕ i ( σ ( t ) ) ) .
(3.13)
We know from (3.6) that there is a t t 9 such that z Δ ( t ) > 0 and z Δ n 1 ( t ) > 0 for an even n. Since z ( t ) > 0 is increasing z ( δ 1 ϕ i ( σ ( t ) ) ) z ( δ 1 ϕ i ( t ) ) for i = 1 , 2 , , m . Therefore, by Lemma 5,
z ( δ 1 ϕ i ( t ) ) λ ( 1 ) n 1 g n 1 ( σ ( ϕ i ( t ) ) , ϕ i ( t ) ) z Δ n 1 ( ϕ i ( t ) ) .
(3.14)
Then by Δ-derivating (3.14) and using g n 1 Δ ( σ ( t ) , t ) = g n 2 σ ( σ ( t ) , t ) , we get
[ z ( δ 1 ϕ i ( t ) ) ] Δ λ ( 1 ) n 2 g n 1 Δ ( σ ( ϕ i ( t ) ) , ϕ i ( t ) ) z Δ n 1 ( ϕ i ( t ) ) λ ( 1 ) n 2 g n 2 σ ( σ ( ϕ i ( t ) ) , ϕ i ( t ) ) z Δ n 1 ( ϕ i ( t ) )
by Lemma 2
z Δ ( δ 1 ϕ i ( t ) ) δ 1 ϕ i Δ ( t ) λ ( 1 ) n 2 g n 2 σ ( σ ( ϕ i ( t ) ) , ϕ i ( t ) ) z Δ n 1 ( ϕ i ( t ) ) .
Since ϕ i ( t ) t , we obtain
z Δ ( δ 1 ϕ i ( t ) ) δ λ ( 1 ) n 2 g n 2 σ ( σ ( ϕ i ( t ) ) , ϕ i ( t ) ) z Δ n 1 ( t ) ϕ i Δ ( t ) .
(3.15)
Hence by (3.10), (3.14) and (3.15), we have
w Δ ( t ) α i = 1 m Q i ( t ) z ( ϕ i ( t ) ) z ( δ 1 ϕ i ( t ) ) δ 1 w σ ( t ) δ λ ( 1 ) n 2 g n 2 σ ( σ ( ϕ i ( t ) ) , ϕ i ( t ) ) z Δ n 1 ( t ) ϕ i Δ ( t ) ϕ i Δ ( t ) z ( δ 1 ϕ i ( σ ( t ) ) ) α i = 1 m Q i ( t ) z ( ϕ i ( t ) ) z ( δ 1 ϕ i ( t ) ) δ 1 w σ ( t ) δ λ ( 1 ) n 2 g n 2 σ ( σ ( ϕ i ( t ) ) , ϕ i ( t ) ) ϕ i Δ ( t ) ϕ i Δ ( t ) z Δ n 1 ( σ ( t ) ) z ( δ 1 ϕ i ( σ ( t ) ) ) α i = 1 m Q i ( t ) λ ( 1 ) n 2 g n 2 σ ( σ ( ϕ i ( t ) ) , ϕ i ( t ) ) ( w σ ( t ) ) 2 ,
and then
α i = 1 m Q i ( t ) w Δ ( t ) λ ( 1 ) n 2 w 2 ( t ) g n 2 σ ( σ ( ϕ i ( t ) ) , ϕ i ( t ) )
(3.16)
for t t 10 . If we multiply (3.16) by φ ( t ) and integrate it from t 10 to t, we obtain by Theorem 2
α t 10 t φ ( s ) i = 1 m Q i ( s ) Δ s t 10 t φ ( s ) w Δ ( s ) Δ s t 10 t λ ( 1 ) n 2 φ ( s ) w 2 ( s ) g n 2 σ ( σ ( ϕ i ( s ) ) , ϕ i ( s ) ) Δ s [ φ ( t ) w ( t ) φ ( t 10 ) w ( t 10 ) t 10 t φ Δ ( s ) w σ ( t ) Δ s ] t 10 t λ ( 1 ) n 2 φ ( s ) w 2 ( s ) g n 2 σ ( σ ( ϕ i ( s ) ) , ϕ i ( s ) ) Δ s φ ( t 10 ) w ( t 10 ) + t 10 t φ Δ ( s ) w σ ( t ) Δ s λ t 10 t φ ( s ) w 2 ( s ) g n 2 σ ( σ ( ϕ i ( s ) ) , ϕ i ( s ) ) Δ s φ ( t 10 ) w ( t 10 ) λ t 10 t φ ( s ) g n 2 σ ( σ ( ϕ i ( s ) ) , ϕ i ( s ) ) × [ w ( s ) φ Δ ( s ) 2 λ φ ( s ) g n 2 σ ( σ ( ϕ i ( s ) ) , ϕ i ( s ) ) ] 2 Δ s + t 10 t [ φ Δ ( s ) ] 2 4 λ φ ( s ) g n 2 σ ( σ ( ϕ i ( s ) ) , ϕ i ( s ) ) Δ s φ ( t 10 ) w ( t 10 ) + t 10 t [ φ Δ ( s ) ] 2 4 λ φ ( s ) g n 2 σ ( σ ( ϕ i ( s ) ) , ϕ i ( s ) ) Δ s .
Therefore by (C3),
= α lim t sup t 10 t φ ( s ) i = 1 m Q i ( s ) Δ s φ ( t 10 ) w ( t 10 ) + 1 4 λ lim t sup t 10 t [ φ Δ ( s ) ] 2 φ ( s ) g n 2 σ ( σ ( ϕ i ( s ) ) , ϕ i ( s ) ) Δ s < .

This is a contradiction.

Now let us consider the case of y ( t ) < 0 for t t 1 . By (1.1) and (1.2), we have
z Δ n ( t ) = i = 1 m Q i ( t ) f i ( y ( ϕ i ( t ) ) ) > 0

for t t 1 . That is, z Δ n > 0 . It follows that z Δ j ( t ) ( j [ 0 , n 1 ] N 0 ) is strictly monotone and eventually of constant sign. Since P ( t ) is an oscillatory function, there exists a t 2 t 1 such that z ( t ) < 0 for t t 2 . Since y ( t ) is bounded, by (C1) and (1.2), there is a t 3 t 2 such that z ( t ) is also bounded for t t 3 . Assume that x ( t ) = z ( t ) . Then x Δ n ( t ) = z Δ n ( t ) . Therefore, x ( t ) > 0 and x Δ n ( t ) < 0 for t t 3 . Hence, we observe that x ( t ) is bounded. Since n is odd, by Lemma 1, there exists a t 4 t 3 and p = 1 (otherwise x ( t ) is not bounded) such that ( 1 ) k x Δ k ( t ) > 0 , k [ 0 , n 1 ] N 0 and t t 4 . That is, ( 1 ) k z Δ k ( t ) < 0 , k [ 0 , n 1 ] N 0 and t t 4 . In particular, for t t 4 we have z Δ ( t ) > 0 . Therefore, z ( t ) is increasing. For the rest of the proof, we can proceed the proof similarly to the case of y ( t ) > 0 . Hence, the proof is completed. □

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Sciences and Arts, Usak University, 1 Eylul Campus, Usak, 64200, Turkey
(2)
Department of Mathematics, Faculty of Sciences and Literatures, Kastamonu University, Kastamonu, Turkey

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